Answer
\[\frac{1}{2}\ln |x^2-x-6|+\frac{3}{10}\ln |x-3|-\frac{3}{10}\ln |x+2|+C\]
Where $C$ is constant of integration
Work Step by Step
Let \[I=\int\frac{x+1}{x^2-x-6}dx\]
\[x+1=A\frac{d}{dx}(x^2-x-6)+B\]
\[x+1=A(2x-1)+B\]
\[x+1=2Ax+(B-A)\]
Compare like coefficients both side
\[2A=1\Rightarrow A=\frac{1}{2}\]
\[B-A=1\Rightarrow B=\frac{3}{2}\]
\[\Rightarrow I=\frac{1}{2}\int\frac{2x-1}{x^2-x-6}dx+\frac{3}{2}\int\frac{dx}{x^2-x-6}\;\;\;...(1)\]
Let \[I_1=\frac{1}{2}\int\frac{2x-1}{x^2-x-6}dx\]
Make a substitution \[t=x^2-x-6\;\;\;\;...(a)\]
\[\Rightarrow dt=(2x-1)dx\]
\[I_1=\frac{1}{2}\int\frac{dt}{t}\]
\[I_1=\frac{1}{2}\ln |t|\]
From ($a$)
\[I_1=\frac{1}{2}\ln |x^2-x-6|\;\;\;\;...(2)\]
Let \[I_2=\frac{3}{2}\int\frac{dx}{x^2-x-6}\]
\[I_2=\frac{3}{2}\int\frac{dx}{x^2-3x+2x-6}\]
\[I_2=\frac{3}{2}\int\frac{dx}{(x-3)(x+2)}\]
We first determine the partial fraction decomposition of the integrand
\[\frac{1}{(x-3)(x+2)}=\frac{A}{(x-3)}+\frac{B}{(x+2)}\]
\[1=A(x+2)+B(x-3)\]
\[1=(A+B)x+(2A-3B)\]
Comparing like coefficients both side
\[A+B=0\Rightarrow B=-A\;\;\;...(b)\]
\[2A-3B=1\]
From ($b$)
\[5A=1\Rightarrow A=\frac{1}{5}\]
From $(b)$
\[B=\frac{-1}{5}\]
\[\frac{1}{(x-3)(x+2)}=\frac{1}{5(x-3)}-\frac{1}{5(x+2)}\]
\[\Rightarrow I_2=\frac{3}{2}\int\left[\frac{1}{5(x-3)}-\frac{1}{5(x+2)}\right]dx\]
\[\Rightarrow I_2=\frac{3}{10}\int\frac{1}{x-3}dx-\frac{3}{10}\int\frac{1}{x+2}dx\]
\[\Rightarrow I_2=\frac{3}{10}\ln |x-3|-\frac{3}{10}\ln |x+2|\;\;\;...(3)\]
Using (2) and (3) in (1)
\[I=\frac{1}{2}\ln |x^2-x-6|+\frac{3}{10}\ln |x-3|-\frac{3}{10}\ln |x+2|+C\]
Where $C$ is constant of integration
\[I=\frac{1}{2}\ln |x^2-x-6|+\frac{3}{10}\ln |x-3|-\frac{3}{10}\ln |x+2|+C\]
